An AP consists of 50 terms of which 3rd term is 12 and the last term is
106. Find the 29th term
Answers
Given:
An AP consists of 50
3rd term is 12 and the last term is 106.
Find:
The 29th term
Solution:
The 3rd term of Ap is 12
The last term of Ap is 106
we know that;
The 3rd term of Ap is 12.
an = a + (n - 1)d
a3 = a + (3 - 1)d
12 = a + 2d .............(i).
The last term of Ap is 106.
=> an = a + (n - 1)d
=> a50 = a + (50 - 1)d
=> 106 = a + 49d .............(ii).
Now, Subtracting Eq. (ii) and (i) we get,
=> y = 94/47
=> y = 2
Now, putting the value of y in Eq. (i).
=> a + 2d = 12
=> a + 2(2) = 12
=> a + 4 = 12
=> a = 12 - 4
=> a = 8
So,
=> a29 = a + (29 - 1)d
=> a29 = a + 28d
=> a29 = 8 + 28 × 2
=> a29 = 8 + 56
=> a29 = 64
Hence the 29th term of Ap is 64.
I hope it will help you.
Regards.
Answer:
an = a + (n - 1)d
a3 = a + (3 - 1)d
12 = a + 2d .............(i).
The last term of Ap is 106.
=> an = a + (n - 1)d
=> a50 = a + (50 - 1)d
=> 106 = a + 49d .............(ii).
Now, Subtracting Eq. (ii) and (i) we get,
=> y = 94/47
=> y = 2
Now, putting the value of y in Eq. (i).
=> a + 2d = 12
=> a + 2(2) = 12
=> a + 4 = 12
=> a = 12 - 4
Step-by-step explanation: